What you did is a very good first step.
For the second integral, make the substitution $u=1-x^2$. We end up with the easy $\int -\frac{1}{2}u^{-3/2}\,du $.

For the first integral, note that the bottom simplifies to $(1-x^2)\sqrt{1-x^2}$, since the square root is only defined when $|x|\le 1$. There is nice cancellation, and we are integrating $\frac{1}{\sqrt{1-x^2}}$, easy, we get an $\arcsin$.